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Problem. Let $G$ be a residually finite group, and identify $G$ with its image under the canonical map to its profinite completion $\hat{G}$. Let $x,y \in G$. Prove that the following conditions are equivalents:

(i) $x,y$ are conjugate in $\hat{G}$.

(ii) the imagens of $x,y$ in $G/K$ are conjugate in $G/K$, for every normal subgroup $K$ of finite index in $G$.

I'm sure that I'm confusing about this problem, so I'd like some help.

We know that $$\hat{G} = \varprojlim_{K \in I}G/K$$ where $I$ is a non-empty filter base of normal subgroups of finite index in $G$ and there is a continuous homomorphism $\varphi: G \to \hat{G}$ given by $\varphi(g) = Kg$. The pair $(\hat{G},\varphi)$ has the property:

"if $\psi: G \to H$ is a continuous homomorphism, to a finite group $H$, then there is a unique homomorphism $\overline{\psi}: \hat{G} \to H$ such that $\psi = \overline{\psi}\circ\varphi$."

  1. If $x,y \in G$, then $Kx,Ky \in \hat{G}$. So, the item (i) must be "... in $G$"?

  2. Who are the imagens of $x,y$ in $G/K$? Since $\hat{G} = \varprojlim G/K$, then $(G/K_{i},f_{ij})$ is a inverse system with inverse limit $(\hat{G},f_{i})$ where $f_{i}: \hat{G} \to G/K_{i}$? We can connect $G$ to $\hat{G}$ by $\varphi$ and $\hat{G}$ to $G/K_{i}$ by $f_{i}$, but who is $f_{i}$? I don't know a definition of completion using, explicitly, inverse systems indexed by a direct set and its maps. Thus, I don't know how to start this problem.

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    $\begingroup$ $\hat{G}$ is the set of sequences $(a_K K)_{K \in I}$ (indexed by finite index normal subgroups $K$) such that $a_K \in G, a_K K\in G/K$ and $a_K H = a_H H$ whenever $K\subset H$. Trying with $G = \Bbb{Z}$ and $I$ the index $p^n$ subgroups yields the $p$-adic integers, with $I$ every subgroup it is the profinite integers. $\endgroup$ – reuns Apr 21 at 20:28
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    $\begingroup$ Just a word on your profile: if you're tackling problems like this, then, surely, you'll be able to help someone. Have confidence. You're doing well :) $\endgroup$ – Shaun Apr 21 at 20:51
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    $\begingroup$ @Shaun thank you for the encouragement! $\endgroup$ – Lucas Corrêa Apr 21 at 21:04
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    $\begingroup$ You can also think to $\hat{G}$ as the set of (limits of) sequences $(g_n)_{n \ge 1}$ such that $g_n K$ ends being constant for every $K$, then $a,b\in G$ are conjugate in $\hat{G}$ if $a^{-1} (g_n b g_n^{-1}) K \to K$ for every $K$. $\endgroup$ – reuns Apr 21 at 21:31
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    $\begingroup$ $p$-adic integers $\Bbb{Z}_p$ are the completion of $\Bbb{Z}$ for the $p$-adic metric $d(b,b+a) = |a|_p = p^{-k}$ if $a \equiv 0 \bmod p^k,a \not \equiv 0 \bmod p^{k+1}$. Do you see how an equivalent metric can be defined in any group $G$ to obtain $\hat{G}$ as the completion ? $\endgroup$ – reuns Apr 21 at 21:50
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You don't need a "filter basis of normal subgroups" to define $\hat{G}$. Take $\hat{G}= \underset{\leftarrow}{\mathrm{lim}} \: G/K$ over every normal subgroup of finite index $K$ of $G$. Then $\hat{G}$ is a subgroup of the product $\prod_K G/K$ (over the same $K$'s) as in the comment by @reuns, and one has projections $p_K :\hat{G} \to G/K$ for every normal subgroup of finite index $K$ of $G$, that are factorizations of the common projections $\pi_K : G \to G/K$ : $\pi_K = G \overset{i}{\to} \hat{G} \overset{p_K}{\to} G/K$ (where $i$ is the natural inclusion).

So $(i) \Rightarrow (ii)$ is pretty clear.

For the other inclusion, you have to build an element $z=(z_K)_K \in \hat{G}$ such that $zxz^{-1}=y$ from $z_K$'s such that $z_K p_K(x) z_K^{-1}=p_K(y)$ for every normal subgroup of finite index $K$ of $G$.

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  • $\begingroup$ I think I got your idea! I will try. $\endgroup$ – Lucas Corrêa Apr 21 at 21:04
  • $\begingroup$ Actually you can keep the definition with a basis, you'll just need to recover any normal subgroup of finite index from the basis. It works the same way $\endgroup$ – AlexL Apr 21 at 21:16

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